题目附件:

通过百度网盘分享的文件：御宛题.zip
链接：https://pan.baidu.com/s/1i15jdms-Min9xmeDqasKkw?pwd=1fge 
提取码：1fge 
--来自百度网盘超级会员V3的分享

MIISC

我敲，黑客

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求出密码
exp:

from Crypto.Util.number import * 
res = 0
for i in range(100,1000):
 if isPrime(i):
    res+=i 
print(res)
# 75067

或者直接爆破密码

解压压缩包

010查看文件

解压后得到jpg图片，在图片的尾部发现一串base64编码

扫描二维码

最后flag为flag{asdf%^&*ghjkl}

List of file signatures

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010查看文件

这个图片每隔4个比特反转一次

exp:

file = open('fl4g.jpeg','rb').read()
res = b''
for i in range(0,len(file),4):
    res += file[i:i+4][::-1]
open('flag.jpg','wb').write(res)

运行得到

或者直接使用b神工具(真神)

查看目录

最后flag为flag{byt3_sw4p}

这能执行吗？

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ida查看得到用户名和密码

发现base编码

提取出来base转图片

最后flag为flag{Arm0uR_pPTi4}

又是二维码捏

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扫描二维码

base64—-rot13

最后flag为flag{n0_body_f0rget_qr_code}

把回忆拼好给你

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提取两个图片的像素点然后拼起来，得到的新图片就是flag
exp:

import cv2

img1 = cv2.imread(r'E:\\huiyi\\1.png',-1)

  

# print(img1[2])

img2 = cv2.imread(r'E:\\huiyi\\2.png',-1)

new_img = []

  

for i in range(len(img1)):

    img=[]

    for j in range(len(img1)):

        if img1[i][j]==img2[i][j]:

            img += [img1[i][j]]

        else:

            img += [max(img1[i][j],img2[i][j])]

    new_img += [img]

# print(new_img)

from PIL import Image

x = len(img1)

y = len(img1)

  

im = Image.new("RGB", (x, y))

  

for i in range(0, x):

    for j in range(0, y):

        im.putpixel((i, j), (new_img[j][i], new_img[j][i], new_img[j][i]))

im.save("E:\\huiyi\\flag.png")

运行得到

或者使用stegsolve双图异或

最后flag为flag{I_L0V3_PYTH0N}

python？re？哦耶

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import base64 as rtfd

import webbrowser

import time

def mikeSwift(cre):

    sto = []

    gre = ""

    for i in cre:

        sto.append(i+str(len(i)))

        sto.append("h4ck" + i)

    for i in sto:

        gre+=i

    return gre

def prompt():

    return bytes(input("Welcome to the loading dock. What is the password?\t"), 'utf-8')

def obfuscate(bys):

    fusc = rtfd.b64encode(bys)

    fusc += b"534345fdfgfgfdhty6y56yjl"

    fusc = str(fusc)

    fusc = fusc[2:len(fusc)-1]

    refus = []

    for i in fusc:

        refus.append((str(i)))

    fusc="florSFIUEfet4565477"

    for i in refus:

        fusc+=i

    return fusc

def crypt(sor):

    sro = []

    fusc = "893"

    for i in range(len(sor)):

        sro.append(sor[i]+str(i))

    sro.reverse()

    for i in sro:

        fusc+=i

    return fusc

def grant():

    print("Congartulation. Pleas Procid")

    webbrowser.open("https://ctflearn.com/index.php?action=find_problem_details&problem_id=449")

def punish():

    print("This is going to hurt.")

    while True:

        time.sleep(.1)

        webbrowser.open("https://www.youtube.com/watch?v=O3asoGVHix8")

def main():

    sik1 = prompt()

    sik = obfuscate(sik1)

    sik = crypt(sik)

    sik = mikeSwift(sik)

    if sik == "81h4ck891h4ck931h4ck3l1h4ckl81h4ck821h4ck2j1h4ckj81h4ck811h4ck1y1h4cky81h4ck801h4ck061h4ck671h4ck791h4ck951h4ck571h4ck781h4ck8y1h4cky71h4ck771h4ck761h4ck671h4ck761h4ck6y1h4cky71h4ck751h4ck5t1h4ckt71h4ck741h4ck4h1h4ckh71h4ck731h4ck3d1h4ckd71h4ck721h4ck2f1h4ckf71h4ck711h4ck1g1h4ckg71h4ck701h4ck0f1h4ckf61h4ck691h4ck9g1h4ckg61h4ck681h4ck8f1h4ckf61h4ck671h4ck7d1h4ckd61h4ck661h4ck6f1h4ckf61h4ck651h4ck551h4ck561h4ck641h4ck441h4ck461h4ck631h4ck331h4ck361h4ck621h4ck241h4ck461h4ck611h4ck131h4ck361h4ck601h4ck051h4ck551h4ck591h4ck9=1h4ck=51h4ck581h4ck801h4ck051h4ck571h4ck7n1h4ckn51h4ck561h4ck6R1h4ckR51h4ck551h4ck5s1h4cks51h4ck541h4ck4R1h4ckR51h4ck531h4ck3z1h4ckz51h4ck521h4ck2Z1h4ckZ51h4ck511h4ck1f1h4ckf51h4ck501h4ck0V1h4ckV41h4ck491h4ck9T1h4ckT41h4ck481h4ck8M1h4ckM41h4ck471h4ck7f1h4ckf41h4ck461h4ck6N1h4ckN41h4ck451h4ck5H1h4ckH41h4ck441h4ck4Z1h4ckZ41h4ck431h4ck3y1h4cky41h4ck421h4ck2R1h4ckR41h4ck411h4ck1z1h4ckz41h4ck401h4ck0d1h4ckd31h4ck391h4ck9r1h4ckr31h4ck381h4ck8N1h4ckN31h4ck371h4ck7G1h4ckG31h4ck361h4ck6N1h4ckN31h4ck351h4ck5i1h4cki31h4ck341h4ck491h4ck931h4ck331h4ck311h4ck131h4ck321h4ck2Z1h4ckZ31h4ck311h4ck101h4ck031h4ck301h4ck0w1h4ckw21h4ck291h4ck9m1h4ckm21h4ck281h4ck8R1h4ckR21h4ck271h4ck771h4ck721h4ck261h4ck6J1h4ckJ21h4ck251h4ck5X1h4ckX21h4ck241h4ck4Z1h4ckZ21h4ck231h4ck3i1h4cki21h4ck221h4ck2l1h4ckl21h4ck211h4ck131h4ck321h4ck201h4ck0Y1h4ckY11h4ck191h4ck971h4ck711h4ck181h4ck871h4ck711h4ck171h4ck741h4ck411h4ck161h4ck651h4ck511h4ck151h4ck561h4ck611h4ck141h4ck451h4ck511h4ck131h4ck341h4ck411h4ck121h4ck2t1h4ckt11h4ck111h4ck1e1h4cke11h4ck101h4ck0f1h4ckf91h4ck9E1h4ckE81h4ck8U1h4ckU71h4ck7I1h4ckI61h4ck6F1h4ckF51h4ck5S1h4ckS41h4ck4r1h4ckr31h4ck3o1h4cko21h4ck2l1h4ckl11h4ck1f1h4ckf01h4ck0":

        grant()

    else:

        punish()

main()

整个程序的逻辑是：base64加密 –> 前后拼上垃圾字符 –> 将字符每一位和下标拼起来组成新字符串放在一个数组中 –> 反转这个数组 –> 将所有字符串拼起来，并在首部添上 893 –> 将字符每一位和下标拼起来，再拼上 h4ck 和这个字符，得到新字符串按照程序逻辑逆就完事了

解密脚本
exp:

import base64

s="81h4ck891h4ck931h4ck3l1h4ckl81h4ck821h4ck2j1h4ckj81h4ck811h4ck1y1h4cky81h4ck801h4ck061h4ck671h4ck791h4ck951h4ck571h4ck781h4ck8y1h4cky71h4ck771h4ck761h4ck671h4ck761h4ck6y1h4cky71h4ck751h4ck5t1h4ckt71h4ck741h4ck4h1h4ckh71h4ck731h4ck3d1h4ckd71h4ck721h4ck2f1h4ckf71h4ck711h4ck1g1h4ckg71h4ck701h4ck0f1h4ckf61h4ck691h4ck9g1h4ckg61h4ck681h4ck8f1h4ckf61h4ck671h4ck7d1h4ckd61h4ck661h4ck6f1h4ckf61h4ck651h4ck551h4ck561h4ck641h4ck441h4ck461h4ck631h4ck331h4ck361h4ck621h4ck241h4ck461h4ck611h4ck131h4ck361h4ck601h4ck051h4ck551h4ck591h4ck9=1h4ck=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"

ss=''

for i in range(0,len(s),7):

    ss+=s[i]

tmp=ss[3:]

sss=''

for i in range(0,220,3):

    sss+=tmp[i]

for i in range(221,len(tmp),2):

    sss+=tmp[i]

sss=sss[::-1]

print(base64.b64decode(sss[19:][:-24]))

运行得到

最后flag为flag{Fl4g_b4ckw4rds_15_g4lF}

把回忆拼好给你2.0

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题目给了500张一细条的图片，显然就是要把这些图片按照图片名上下拼接起来
exp:

from PIL import Image

def vertical_concatenate_images(image_paths, output_path):

    first_image = Image.open(image_paths[0])

    width, height = first_image.size

    total_height = sum([Image.open(img).size[1] for img in image_paths])

    concatenated_image = Image.new('RGB', (width, total_height))

    y_offset = 0

    for img_path in image_paths:

        img = Image.open(img_path)

        concatenated_image.paste(img, (0, y_offset))

        y_offset += img.size[1]

        concatenated_image.save(output_path)

image_paths = [f'{i}.png' for i in range(0, 500)]

output_path = 'concatenated_image.jpg'

vertical_concatenate_images(image_paths, output_path)

运行得到

hex解密

最后flag为flag{th3_KGB_l0v3s_CTF}

不会真有人一个一个解压缩吧？

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压缩包套娃
调用系统命令用unzip解压
exp:

import os

import shutil

def unzip(zipname,pwd,filePath2):

    print(zipname,pwd,filePath2)

    os.system(f'unzip -P {pwd} {zipname} -d {filePath2}')

cnt = 0

while True:

   filePath1 = './'+str(cnt) + '/'

   cnt = (cnt + 1) & 1

   filePath2 = './'+str(cnt) + '/'

  

   for i,j,k in os.walk(filePath1):

       pwd = ''

       if 'password.txt' in k:

            pwd = open(filePath1 + 'password.txt','r').read().strip()

            # print(pwd)

       else:

            break

       for i in k:

           if '.zip' in i:

               # print(i,pwd,filePath2)

               shutil.rmtree(filePath2)

               os.mkdir(filePath2)

               unzip(filePath1+i,pwd,filePath2)

最后flag为flag{1_TH1NK_U_G00D_4T_SCR1PT1N9_botanbell_1s_h3r3^_^}

来签个到吧，包简单的

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stegsolve一把梭

reverse—-base64

最后flag为flag{m3Et_me_4t_12_aM}

中秋特辑（1）

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图片转base64编码

把base64后面的== 改为 !=

md5加密

最后flag为flag{CB08E2546A6167B24C5E490681647A00}

这是？配置文件？

下载文件

MobaXterm 的配置文件，用脚本解密ssh密码

python MobaXtermCipher.py dec -p flag_is_here DLulatnJIPtEF/EMGfysL2F58R4dfQIbQhzwuNqL

base解密

最后flag为flag{you_are_a_g00d_guy}

图片的隐藏

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爆破密码

解压压缩包

宽高一把梭

扫描二维码

最后flag为flag{abcdefghiklmn}

快来社我_1

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百度识图

最后flag为flag{yuntaishan}

快来社我_2

搜索关键词看到这篇文章

最后flag为flag{Sycamore}

快来社我_3

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谷歌识图，找到图片来自这篇文章

最后flag为flag{Carbon_County}

这好熟悉，有点像某个数列

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base64解密

斐波那契数列
每个数字对应的项数转ascii
exp:

from Crypto.Util.number import *

ls=[927372692193078999176,16641027750620563662096,83621143489848422977,1500520536206896083277,22698374052006863956975682,927372692193078999176,7778742049,135301852344706746049,4807526976,43566776258854844738105,32951280099,218922995834555169026,2427893228399975082453,4807526976,59425114757512643212875125]

def find_fibonacci_index(number):

 if number < 0:

     return -1

  

 a, b = 0, 1

 index = 1

  
  

 while b < number:

     a, b = b, a + b

     index += 1

  

 if b == number:

     return index

 else:

     return -1

ans=[]

for i in ls:

    position = find_fibonacci_index(i)

    if position != -1:

         ans.append(position)

print(''.join(chr(i) for i in ans))

运行得到

最后flag为flag{f1b0n4ch0}

好奇怪的图像

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图片只有一像素高，但宽度很离谱，猜测是把图片分成1像素的条然后拼到一起了，写个脚本还原
exp:

import cv2

from PIL import Image

  

img1 = cv2.imread(r'E:\\qiguai\\test.jpg',-1)

# print(img1)

print(img1.shape)

new_img = []

  

for i in range(0,len(img1[0]),23*2*2):

    img = img1[0][i:i+23*2*2]

    new_img += [img]

  

 # print(len(new_img))

  

x = len(new_img)

y = len(new_img[0])

  

im = Image.new("RGB", (x, y))

  

for i in range(0, x):

    for j in range(0, y):

        im.putpixel((i, j), (new_img[i][j][0], new_img[i][j][1], new_img[i]

[j][2]))

im.save("E:\\qiguai\\flag.png")